非常套路 & 码农的一道树套树题……
正解被分块吊起来打……
树状数组套平衡树，平衡树维护个数和权值和。
然后就是套路。
十分艰难地提交了并且心态崩了。
成功超时。
此代码无法在 LibreOJ 上通过。
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using namespace std;
const int BUFF_SIZE = 1 << 20;
char BUFF[BUFF_SIZE],*BB,*BE;
template<class T>
inline void read(T &x)
{
x = 0;
char ch = 0,w = 0;
while(ch < '0' || ch > '9')
w |= ch == '-',ch = gc();
while(ch >= '0' && ch <= '9')
x = (x << 3) + (x << 1) + (ch ^ '0'),ch = gc();
w ? x = -x : x;
}
const int N = 5e4;
const long long mod = 1e9 + 7;
int n,m,a[N + 10];
long long v[N + 10],ans;
struct node
{
int key,rnd,sz;
long long val,sum;
node *lson,*rson;
} *rt[N + 10];
inline node *new_node(int k,long long v)
{
node *ret = new node();
ret->key = k;
ret->val = ret->sum = v;
ret->rnd = rand();
ret->sz = 1;
return ret;
}
inline void up(node *p)
{
p->sz = 1,p->sum = p->val;
if(ls(p))
p->sz += ls(p)->sz,p->sum = (p->sum + ls(p)->sum) % mod;
if(rs(p))
p->sz += rs(p)->sz,p->sum = (p->sum + rs(p)->sum) % mod;
}
node *merge(node *x,node *y)
{
if(!x)
return y;
if(!y)
return x;
if(x->rnd < y->rnd)
{
rs(x) = merge(rs(x),y);
up(x);
return x;
}
else
{
ls(y) = merge(x,ls(y));
up(y);
return y;
}
}
void split(node *p,int k,node *&x,node *&y)
{
if(!p)
{
x = y = NULL;
return ;
}
if(!k)
{
x = NULL,y = p;
return ;
}
if(k == n)
{
x = p,y = NULL;
return ;
}
if(p->key <= k)
x = p,split(rs(p),k,rs(p),y);
else
y = p,split(ls(p),k,x,ls(p));
up(p);
}
void update(int x,int key,long long v,int k)
{
node *a,*b,*c;
for(;x <= n;x += lowbit(x))
if(k)
{
split(rt[x],key,a,b);
rt[x] = merge(merge(a,new_node(key,v)),b);
}
else
{
split(rt[x],key,a,c);
split(a,key - 1,a,b);
if(b)
delete b;
rt[x] = merge(a,c);
}
}
long long query(int x,int keyl,int keyr)
{
long long ret = 0;
node *a,*b,*c;
for(;x;x -= lowbit(x))
{
split(rt[x],keyr,a,c);
split(a,keyl - 1,a,b);
if(b)
ret = (ret + b->sum) % mod;
rt[x] = merge(merge(a,b),c);
}
return ret;
}
int ask(int x,int keyl,int keyr)
{
int ret = 0;
node *a,*b,*c;
for(;x;x -= lowbit(x))
{
split(rt[x],keyr,a,c);
split(a,keyl - 1,a,b);
if(b)
ret += b->sz;
rt[x] = merge(merge(a,b),c);
}
return ret;
}
int main()
{
read(n),read(m);
int x;
long long val;
for(register int i = 1;i <= n;++i)
{
read(x);
read(val);
a[i] = x,v[i] = val;
update(i,x,val,1);
}
for(register int i = 1;i <= n;++i)
ans = (ans + v[i] * (ask(n,1,a[i] - 1) - ask(i,1,a[i] - 1)) + (query(n,1,a[i] - 1) - query(i,1,a[i] - 1))) % mod;
int y;
while(m--)
{
read(x),read(y);
if(x == y)
{
printf("%lld\n",ans);
continue;
}
if(x > y)
swap(x,y);
ans = (ans - v[x] * (ask(y,1,a[x] - 1) - ask(x,1,a[x] - 1)) - (query(y,1,a[x] - 1) - query(x,1,a[x] - 1))) % mod;
ans = (ans - v[y] * (ask(y - 1,a[y] + 1,n) - ask(x - 1,a[y] + 1,n)) - (query(y - 1,a[y] + 1,n) - query(x - 1,a[y] + 1,n))) % mod;
if(a[x] > a[y])
ans = (ans + v[x] + v[y]) % mod;
update(x,a[x],v[x],0),update(y,a[y],v[y],0);
swap(a[x],a[y]),swap(v[x],v[y]);
update(x,a[x],v[x],1),update(y,a[y],v[y],1);
ans = (ans + v[x] * (ask(y,1,a[x] - 1) - ask(x,1,a[x] - 1)) + (query(y,1,a[x] - 1) - query(x,1,a[x] - 1))) % mod;
ans = (ans + v[y] * (ask(y - 1,a[y] + 1,n) - ask(x - 1,a[y] + 1,n)) + (query(y - 1,a[y] + 1,n) - query(x - 1,a[y] + 1,n))) % mod;
if(a[x] > a[y])
ans = (ans - v[x] - v[y]) % mod;
ans = (ans + mod) % mod;
printf("%lld\n",ans);
}
}